\beginsection 18.6

Prove that $x=\cos x$ for some $x$ in $(0,\pi/2)$.

\medskip

Solution: We have $\cos(0)=1$ and $\cos(\pi/2)=0$.
The problem is that the interval is open, not closed as required by IVT.
However, since $\cos(0)\ne0$ and $\cos(\pi/2)\ne\pi/2$ we know that
$\cos x=x$ does not occur at the endpoints.
Therefore it is safe to use the closed interval $[0,\pi/2]$.
%(Is this o.k. or b.s.?)
Use the trick from Example 1: $f(x)=x-\cos x$.
We have $f(0)=0-1=-1$ and $f(\pi/2)=\pi/2-0=\pi/2$ and
therefore $f(0)<0<f(\pi/2)$.
By IVT there exits an $x_0$ such that $f(x_0)=0$.
Because $f(x_0)=x_0-\cos x_0=0$, we have a point $x_0$
at which $x=\cos x$.
We have already shown that $x\ne\cos x$ at either $0$ or $\pi/2$ so
$x=\cos x$ must occur on the interval $(0,\pi/2)$.

